| Find the equation of the normal to the curve $ \displaystyle y = 1 - x^2$ at the point where the curve crosses the positive $ \displaystyle x$-axis. Find also the coordinates of the point where the normal meets the curve again. |
$ \displaystyle y = 1 - x^2$ αိုαဲ့ curve α α‘ေαါα္း $ \displaystyle x$ αα္αိုးαို ျαα္αြားαဲ့ ေααာαွာ αွိαဲ့ normal line equation αို αွာေαးαါ။ ၎ normal line α curve αို ေαာα္ αα ္ααိα္ ျαα္αြားαဲ့ α‘αွα္αိုαα္း αွာေαးαါ။
Curve α $ \displaystyle x$ αα္αိုးαို $ \displaystyle y=0$ ျαα
္αဲ့ ေααာαွာ ျαα္αာေαါ့။ αါαိုαα္ ေαးαားαဲ့ curve : $ \displaystyle 1 - x^2$ αို 0 αဲα αီေαးαိုα္αα္ curve ျαα္αြားαဲ့ $ \displaystyle x$ αα္αိုးေααα α‘αွα္ေαြ ααΏαီေαါ့။ α‘ေαါα္း $ \displaystyle x$ αα္αိုးαိုα ေျαာαားαာαိုα $ \displaystyle x$ $ \displaystyle \text{coordinate}$ αွာ α‘ႏႈα္αါαာαα္ α‘ႏႈα္ αα္αိုးαို αα္ααါαα္။
Normal Line αဲ့ equation αို αွာαိုα gradient (slope) αိုαα္း αိααα္။ Normal Line αိုαာ tangent ေαα ေαာα့္αα္α်αိုα tangent gradient αဲ့ α‘ႏႈα္αွα္αိα္း (negative reciprocal) ျαα
္αါαα္။ gradient of tangent =$ \displaystyle m$ ျαα
္αα္αိုαα္ normal αဲ့ gradient α $ \displaystyle -\frac{1}{m}$ ျαα
္αွာေαါ့။ αါαိုαα္ tangent αဲ့ gradient αို αိαα္ normal line equation αို $ \displaystyle (y-{{y}_{1}})=-\frac{1}{m}(x-{{x}_{1}})$ αိုαာαဲα αွာαိုα္αα္ ααΏαီ။
Gradient of tangent = $ \displaystyle \frac{dy}{dx}$ αိုαာ differentiation from the first principles ααα္းα αိαΏαီးျαα
္αွာαါ။ tangent αဲ့ gradient αို ααိုα ေαးαားαဲ့ curve αို differentiate αုα္αα္ ααါαΏαီ။
Curve αို differentiate αုα္αိုαααဲ့ tangent αဲ့ gradient αိုαာ general form (curve ေαααွာαွိαဲ့ αα္αα့္ေααာαို ααို αိုαိုαα္)။ ေαးαြα္းα ေαးαားαာα α‘ေαါα္း $ \displaystyle x$ αα္αိုးαို ျαα္ေαာေααာαိုα αိုαားαိုα αွာαားαဲ့ $ \displaystyle x$ αα္αိုး ျαα္αွα္ [ $ \displaystyle (x_{1},y_{1})$ αိုα αိုαΎααါαိုα] α‘α ားαြα္းαိုα္αွ αိုα်α္αဲ့ tangent line αဲ့ gradient αို ααွာေαါ့။ αိုαိုαာα $ \displaystyle m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}$ αါ။
Normal Line equation ααα္ ေαးαြα္းα αα္αα့္ေαးαားαာα ၎ Normal Line α Curve αို ေαာα္αα ္ααိα္ ျαα္αဲ့α‘αွα္αို αွာေαးαါαိုα αိုαားαါαα္။
Line ႏွα ္ေαΎαာα္းျαα္αွα္αိုαာ α‘αα္αα္း α‘αα့္αွာααα္းα αိαဲ့αΏαီးαါαΏαီ။ ၎ Line ႏွα ္ေαΎαာα္းαို ααΏαိဳα္αα္ ေျααα္ေα αဲ့ $ \displaystyle x$ - $ \displaystyle \text{coordinate}$ αဲα $ \displaystyle y$ - $ \displaystyle \text{coordinate}$ αို αွာαိုα္းαာ ျαα ္αါαα္။
αိုαα္းαူ αါαဲ။ Curve αဲα Normal Line αိုαျαα္αွα္αိုαာ ၎ Equation ႏွα ္ေαΎαာα္းαို ααΏαိဳα္αα္ ေျααα္ေα αဲ့ $ \displaystyle x$ - $ \displaystyle \text{coordinate}$ αဲα $ \displaystyle y$ - $ \displaystyle \text{coordinate}$ αို αွာေαးααွာ ျαα ္αါαα္။ αါαိုαα္ ေျααွα္းααα့္ α‘αα့္ေαြαို αိေαာα္αΏαီαိုα αα္αါαα္။ αြα္αΎαα့္αΎαα ိုα။
Solution
$ \displaystyle \begin{array}{l}\text{Curve : }y=1-{{x}^{2}}\\\\\text{When the curve cuts the }x-\text{axis, }y=0.\\\\\therefore 1-{{x}^{2}}=0\Rightarrow {{x}^{2}}=1\Rightarrow x=\pm 1\\\\\text{But the required points lies on the }\\\text{positive }x-\text{axis, }\\\\\therefore x=1\\\\\text{Let }({{x}_{1}},{{y}_{1}})=(1,0)\\\\y=1-{{x}^{2}}\Rightarrow \frac{{dy}}{{dx}}=-2x\\\\\therefore m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(1,0)}}}=-2(1)=-2\\\\\text{Hence the equation of normal line at }({{x}_{1}},{{y}_{1}})\ \text{is}\\\\(y-{{y}_{1}})=-\frac{1}{m}(x-{{x}_{1}})\\\\y-0=\frac{1}{2}(x-1)\Rightarrow y=\frac{1}{2}(x-1)\\\\\text{When this normal line meets the curve,}\\\\\frac{1}{2}(x-1)=1-{{x}^{2}}\Rightarrow 2{{x}^{2}}+x-3=0\\\\\therefore (2x+3)(x-1)=0\\\\\therefore x=-\frac{3}{2}\ \text{or}\ x=1\\\\\text{When }x=-\frac{3}{2}\ ,y=\frac{1}{2}(-\frac{3}{2}-1)=-\frac{5}{4}.\end{array}$
$ \displaystyle \text{Therefore, the normal line meets}$
$ \displaystyle \text{the curve again at the point (}-\frac{3}{2},-\frac{5}{4}).$
Curve αို differentiate αုα္αိုαααဲ့ tangent αဲ့ gradient αိုαာ general form (curve ေαααွာαွိαဲ့ αα္αα့္ေααာαို ααို αိုαိုαα္)။ ေαးαြα္းα ေαးαားαာα α‘ေαါα္း $ \displaystyle x$ αα္αိုးαို ျαα္ေαာေααာαိုα αိုαားαိုα αွာαားαဲ့ $ \displaystyle x$ αα္αိုး ျαα္αွα္ [ $ \displaystyle (x_{1},y_{1})$ αိုα αိုαΎααါαိုα] α‘α ားαြα္းαိုα္αွ αိုα်α္αဲ့ tangent line αဲ့ gradient αို ααွာေαါ့။ αိုαိုαာα $ \displaystyle m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}$ αါ။
Normal Line equation ααα္ ေαးαြα္းα αα္αα့္ေαးαားαာα ၎ Normal Line α Curve αို ေαာα္αα ္ααိα္ ျαα္αဲ့α‘αွα္αို αွာေαးαါαိုα αိုαားαါαα္။
Line ႏွα ္ေαΎαာα္းျαα္αွα္αိုαာ α‘αα္αα္း α‘αα့္αွာααα္းα αိαဲ့αΏαီးαါαΏαီ။ ၎ Line ႏွα ္ေαΎαာα္းαို ααΏαိဳα္αα္ ေျααα္ေα αဲ့ $ \displaystyle x$ - $ \displaystyle \text{coordinate}$ αဲα $ \displaystyle y$ - $ \displaystyle \text{coordinate}$ αို αွာαိုα္းαာ ျαα ္αါαα္။
αိုαα္းαူ αါαဲ။ Curve αဲα Normal Line αိုαျαα္αွα္αိုαာ ၎ Equation ႏွα ္ေαΎαာα္းαို ααΏαိဳα္αα္ ေျααα္ေα αဲ့ $ \displaystyle x$ - $ \displaystyle \text{coordinate}$ αဲα $ \displaystyle y$ - $ \displaystyle \text{coordinate}$ αို αွာေαးααွာ ျαα ္αါαα္။ αါαိုαα္ ေျααွα္းααα့္ α‘αα့္ေαြαို αိေαာα္αΏαီαိုα αα္αါαα္။ αြα္αΎαα့္αΎαα ိုα။
Solution
$ \displaystyle \text{Therefore, the normal line meets}$
$ \displaystyle \text{the curve again at the point (}-\frac{3}{2},-\frac{5}{4}).$
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